Answer
a) 1 - lnx - C + lnx + C = 1 = RHS, so y is a solution of the differential equation
b) attached in picture
c) y= $\frac{lnx+2}{x}$
d)y=$\frac{lnx+2-ln2}{x}$
Work Step by Step
a) y= $\frac{lnx+C}{x}$ --> y'= $\frac{x*(\frac{1}{x})-(lnx+C)}{x^2}$ = $\frac{1-lnx-C}{x^2}$
LHS = ${x^2}$y' + xy = ${x^2}$ * $\frac{1-lnx-C}{x^2}$ + x*$\frac{lnx+C}{x}$
=1 - lnx - C + lnx + C = 1 = RHS, so y is a solution of the differential equation
b) A few notes about the graph y = (lnx+C)/x:
1) there is a vertical asymptote of x=0
2)There is a horizontal asymptote of y=0
3) y=0 --> lnx+C = 0 --> x= e^(-C), so there is an x-intercept at e^(-C)
4) y'=0 --> lnx=1-C --> x=e^(1-C), so there's a local maximum at x=e^(1-C)
c) y(1) = 2 --> 2=$\frac{ln1+C}{1}$ --> 2=C, so the solution is y= $\frac{lnx+2}{x}$ (shown in b)
d) y(2) = 1 --> 1=$\frac{ln2+C}{2}$ --> 2+ln2+C --. C=2-ln2, so the solution is y=$\frac{lnx+2-ln2}{x}$ (shown in b)
